How do you solve using the completing the square method $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ ?

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How do you solve using the completing the square method $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ ?

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Answer 1 · 2016-08-06T07:31:43

$x = - 1 - \sqrt{6}$ $x=-1-sqrt6$ or $x = - 1 + \sqrt{6}$ $x=-1+sqrt6$

Explanation:

$x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$

Now, recalling the identity ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2+2ax+a^2$ and comparing it with $x^{2} + 2 x$ $x^2+2x$ , we need to add and subtract ${(\frac{2}{1})}^{2}$ $(2/1)^2$ to complete square. Hence $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ is

$x^{2} + 2 x + 1 - 1 - 5 = 0$ $x^2+2x+1-1-5=0$ or

$(x^{2} + 2 x + 1) - 6 = 0$ $(x^2+2x+1)-6=0$ or

${(x + 1)}^{2} - {(\sqrt{6})}^{2}$ $(x+1)^2-(sqrt6)^2$ or

$(x + 1 + \sqrt{6}) (x + 1 - \sqrt{6}) = 0$ $(x+1+sqrt6)(x+1-sqrt6)=0$

i.e. $x = - 1 - \sqrt{6}$ $x=-1-sqrt6$ or $x = - 1 + \sqrt{6}$ $x=-1+sqrt6$

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How do you solve using the completing the square method $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ ?

1 Answer

Explanation:

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How do you solve using the completing the square method x^2+2x-5=0x2+2x−5=0x^2+2x-5=0?

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How do you solve using the completing the square method $x^{2} + 2 x - 5 = 0$ $x^2+2x-5=0$ ?